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Proof step 1 from proof eq-one (conclusion)

# str = 1, # str0 = 1 ⊦ (* str) = (* str0) + str1 ↔ str = str0 ∧ str1 = ""

Special step normalize

And obtained the following new goals

Proof step 2:
# str = 1, # str0 = 1 ⊦ str = str0 + str1 ↔ str = str0 ∧ str1 = ""